A chemist is making a solution that is to be 55% chlorine. He has one solution that is 40% chlorine and another that is 65% chlorine. He wants 100 liters of the final solution. How much of each should he mix?

1 Answer
Aug 11, 2015

Answer:

You need to mix 40 L of 40% solution and 60 L of 65% solution.

Explanation:

You know that the volume of the final solution is 100 L. If #x# represents the volume of the 40% solution and #y# represents the volume of the 65% solution, then you can say that

#x + y = 100#

Now you need to focus on the chlorine content. Since the first solution is 40% chlorine, then for #x# liters you would get

#(x)""cancel("L solution") * "40 g chlorine"/(100cancel("L solution")) = (0.4 * x)" g chlorine"#

The same can be said for the 65% solution

#(y)""cancel("L solution") * "65 g chlorine"/(100cancel("L solution")) = (0.65 * y)" g chlorine"#

The final solution has a percent concentration of 55% chlorine, which means that it must contain

#100""cancel("L solution") * "55 g chlorine"/(100cancel("L solution")) = "55 g chlorine"#

The second equation you can write will thus relate the amount of chlorine each solution contributes to the final solution

#0.4x + 0.65y = 55#

This is your system of equations

#{(x+y = 100), (0.4x + 0.65y = 55) :}#

Use the first one to write #x# as a function of #y#

#x = 100 - y#

Now use this expression in the second equation to get the value of #y#

#0.4 * (100 - y) + 0.65y = 55#

#40 - 0.4y + 0.65y = 55#

#0.25y = 15 implies y = 15/0.25 = color(green)(60)#

This means that #x# is equal to

#x = 100 - 60 = color(green)(40)#

So, if you mix 40 L of a 40% chlorine solution with 60 L of a 65% chlorine solution you will get 100 L of a 55% chlorine solution.